Let $V$ be an irreducible variety of finite type over a field $k$, $V_{0}\subseteq V$ open and nonempty (or at least dense). Why is \begin{equation*}\operatorname{dim}_{\operatorname{Krull}}(V_{0})=\operatorname{dim}_{\operatorname{Krull}}(V)\text{?}\end{equation*}
My professor uses this result (without mentioning a proof) as if it were somehow self-evident. I have tried to prove it, but I am at a loss. Who can help?
The general statement is NOT true for arbitrary irreducible topological spaces, not even for varieties of finite type over an arbitrary integral domain, see
this question for instance.
Best Answer
This follows from the fact that the dimension of an irreducible variety is equal to the transcendence degree of its function field (over the base field $k$). Since $V_0$ is dense in $V$, it has the same function field.